# How do you find the critical points of #g'(x)=3x^2-6x^2#?

I can't tell you what the coordinate of your critical point is for that equation because I don't know what your original function is.

Remember, the critical value is at a point in the graph where there is a min/max.

I can't tell you what the coordinate of your critical point is for that equation because I don't know what your original function is.

By signing up, you agree to our Terms of Service and Privacy Policy

To find the critical points of ( g'(x) = 3x^2 - 6x^2 ), you first need to find the values of ( x ) where the derivative equals zero or is undefined.

( g'(x) = 3x^2 - 6x^2 ) simplifies to ( g'(x) = -3x^2 ).

Set ( g'(x) ) equal to zero and solve for ( x ): [ -3x^2 = 0 ]

This gives ( x = 0 ) as the only solution.

So, the critical point of ( g'(x) ) is ( x = 0 ).

By signing up, you agree to our Terms of Service and Privacy Policy

- What are the extrema and saddle points of #f(x,y) = x^2y-y^2x#?
- What are the critical values of #f(x)=x-xsqrt(e^x#?
- How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for # f(x) = sqrt(x^2+1) #?
- How do you find the critical values for #f(x)=x^(2/3)+x^(-1/3)#?
- How do you find the x coordinates of the turning points of the function?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7